LeetCode OJ Minimum Window Substring

Given a string S and a string T, find the minimum window in S which will contain all the characters in T in complexity O(n).

For example,

S =

"ADOBECODEBANC"

T =

"ABC"

Minimum window is

"BANC"

.

Note:

If there is no such window in S that covers all characters in T, return the emtpy string

""

.

If there are multiple such windows, you are guaranteed that there will always be only one unique minimum window in S.

char *minWindow(char *S, char *T) {
	int SL = strlen(S), TL = strlen(T);
	bool inT['z' + 1] = { false };
	int num['z' + 1] = { 0 };
	for (int i = 0; i < TL; i++) num[T[i]] += inT[T[i]] = true;
	int ps1 = 0, ps2 = -1, ansP = 0, ansL = SL + 1;
	while (ps2 < SL)
		if (!TL) {
			if (ansL > ps2 - ps1 + 1) ansL = ps2 + 1 - (ansP = ps1);
			if (inT[S[ps1]] && ++num[S[ps1]] > 0) TL++;
			ps1++;
		} else if (--num[S[++ps2]] >= 0 && inT[S[ps2]]) TL--;
	if ((S + ansP)[ansL]) (S + ansP)[ansL] = '\0';
	return ansL == SL + 1 ? "" : S + ansP;
}